Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int N=sc.nextInt();
		double[] dp=new double[N+1];
		for(int i=0; i<=N; i++) {
			dp[i]=0;
		}
		dp[0]=1;
		double[] kakuritu=new double[N];
		for(int i=0; i<N; i++) {
			kakuritu[i]=sc.nextDouble();
		}
		for(int i=0; i<N; i++) {
			for(int j=i; j>=0; j--) {
				dp[j+1]+=dp[j]*kakuritu[i];
				dp[j]*=(1.000d-kakuritu[i]);
			}
		}
		double sum=0;
		for(int i=(N+1)/2; i<=N; i++) {
			sum+=dp[i];
		}
		System.out.println(sum);
	}
}